How do you use partial fractions to find the integral $int (x^3-x+3)/(x^2+x-2)dx$ ?

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Answer 1 · 2016-12-02T08:26:59

THe answer is $=x^2/2-x+ln(∣x-1∣)+ln(∣x+2∣)+C$

The denominator is

$x^2+x-2=(x-1)(x+2)$

But before, let's do a long division

$color(white)(aaaa)$ $x^3-x+3$ $color(white)(aaaa)$ $∣$ $x^2+x-2$

$color(white)(aaaa)$ $x^3+x^2-2x$ $color(white)(aa)$ $∣$ $x-1$

$color(white)(aaaa)$ $0-x^2+x+3$

$color(white)(aaaaaa)$ $-x^2-x+2$

$color(white)(aaaaaaaa)$ $0+2x+1$

So,

$(x^3-x+3)/(x^2+x-2)=x-1+(2x+1)/(x^2+x-2)$

Now we do the decomposition in partial fractions

$(2x+1)/(x^2+x-2)=(2x+1)/((x-1)(x+2))$

$=A/(x-1)+B/(x+2)=(A(x+2)+B(x-1))/((x-1)(x+2))$

so,

$2x+1=A(x+2)+B(x-1))$

Let $x=1$ , $=>$ , $3=3A$ , $=>$ , $A=1$

Let $x=-2$ , $=>$ , $-3=-3B$ , $=>$ , $B=1$

So,

$(x^3-x+3)/(x^2+x-2)=(x-1)+1/(x-1)+1/(x+2)$

$int((x^3-x+3)dx)/(x^2+x-2)=int(x-1)dx+intdx/(x-1)+intdx/(x+2)$

$=x^2/2-x+ln(∣x-1∣)+ln(∣x+2∣)+C$

How do you use partial fractions to find the integral int (x^3-x+3)/(x^2+x-2)dxint (x^3-x+3)/(x^2+x-2)dx?